﻿function ListNode(val,next){
    thisl.val=(val===undefined? 0: val)
    this.next = (next===undefined? null: next)
}

// 方法一：递归
var mergeTwoLists  = function(l1,l2){
    if(l1 === null){
        return l2;
    }else if(l2 ===null) {
        return l1;
    }else if(l1.val < l2.val){
        l1.next = mergeTwoLists(l1.next,l2);
        return l1;
    }else{
        l2.next = mergeTwoLists(l1,l2.next);
        return l2;
    }
};


// 方法二：迭代
var mergeTwoList2 = function(l1,l2){    
    const prehead = new ListNode(-1);
    
    let prev  = prehead;
    while(l1 != null && l2!=null){
        if(l1.val<= l2.val){
            prev.next = l1;
            l1 = l1.next;
        }else{
            prev.next = l2;
            l2= l2.next;
        }
        prev = prev.next;
    }
    
    // 合并后l1和l2最多只有一个还未被合并完，直接将链表末尾指向未合并完的链表即可
    prev.next = l1 === null ? l2 : l1;
    return prehead.next;
};
